[proofplan] We use the singular-value decomposition to write $T-T_n$ as the diagonal tail of $T$ after the first $n$ singular directions. Orthogonality of the vectors $u_j$ gives the upper bound on the tail norm, while testing the tail on $v_{n+1}$ gives the matching lower bound when this vector exists. For the optimality statement, any rank-at-most-$n$ operator must vanish on some non-zero vector in the $(n+1)$-dimensional space spanned by $v_1,\dots,v_{n+1}$; the singular-value ordering then forces $T$ to have norm at least $s_{n+1}$ on that vector. [/proofplan]

[step:Express the compact operator by its singular-value expansion] By the singular-value decomposition theorem for compact operators, [citetheorem:8402], there are an index set $J=\{1,\dots,m\}$ for some $m\in\mathbb N$ or $J=\mathbb N$, positive singular values $(s_j)_{j\in J}$ ordered non-increasingly, an orthonormal set $(v_j)_{j\in J}$ in $(\ker T)^\perp\subset H$, and an orthonormal set $(u_j)_{j\in J}$ in $\overline{\operatorname{Range}(T)}\subset K$ such that, for every $x\in H$. Here $\ker T$ denotes the kernel of $T$, and $\operatorname{Range}(T)$ denotes the range of $T$. \begin{align*} Tx=\sum_{j\in J}s_j(x,v_j)_H u_j. \end{align*} The convergence is in the norm of $K$. The Schmidt truncation is therefore the finite-rank operator \begin{align*} T_nx=\sum_{\substack{j\in J, j\le n}}s_j(x,v_j)_H u_j. \end{align*} Hence, for every $x\in H$, \begin{align*} (T-T_n)x=\sum_{\substack{j\in J, j>n}}s_j(x,v_j)_H u_j. \end{align*} If $J$ has at most $n$ elements, this last sum is empty, so $T-T_n=0$ and $s_{n+1}=0$ by convention. [/step]

[step:Compute the norm of the omitted singular-value tail]Assume first that $J$ contains the index $n+1$. For $x\in H$, the orthonormality of $(u_j)_{j\in J}$ and Parseval's inequality for the orthonormal set $(v_j)_{j\in J}$ give \begin{align*} \|(T-T_n)x\|_K^2 = \sum_{\substack{j\in J, j>n}}s_j^2 |(x,v_j)_H|^2. \end{align*} Since the singular values are non-increasing, $s_j\le s_{n+1}$ for every $j\in J$ with $j>n$. Therefore \begin{align*} \|(T-T_n)x\|_K^2 \le s_{n+1}^2\sum_{\substack{j\in J, j>n}} |(x,v_j)_H|^2 \le s_{n+1}^2\|x\|_H^2. \end{align*} Taking the supremum over all $x\in H$ with $\|x\|_H\le 1$ gives \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)}\le s_{n+1}. \end{align*} For the reverse inequality, evaluate the tail at the unit vector $v_{n+1}$. Since $(v_{n+1},v_j)_H=0$ for $j\ne n+1$ and $(v_{n+1},v_{n+1})_H=1$, \begin{align*} (T-T_n)v_{n+1}=s_{n+1}u_{n+1}. \end{align*} Because $\|u_{n+1}\|_K=1$, this gives \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)} \ge \|(T-T_n)v_{n+1}\|_K = s_{n+1}. \end{align*} Thus \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)}=s_{n+1}. \end{align*} If $J$ has at most $n$ elements, the previous step already showed $T-T_n=0$ and $s_{n+1}=0$, so the same equality holds.[/step]

[guided]We now compute the operator norm of the error made by discarding all singular directions after the first $n$. Assume first that the singular value $s_{n+1}$ exists, meaning that $J$ contains $n+1$. For each $x\in H$, the singular-value expansion gives \begin{align*} (T-T_n)x=\sum_{\substack{j\in J, j>n}}s_j(x,v_j)_H u_j. \end{align*} The point of writing the error this way is that the vectors $u_j$ are orthonormal in $K$. Therefore the norm squared of the sum is the sum of the squared coefficients: \begin{align*} \|(T-T_n)x\|_K^2 = \sum_{\substack{j\in J, j>n}}s_j^2 |(x,v_j)_H|^2. \end{align*} The ordering $s_1\ge s_2\ge\cdots$ now gives the decisive estimate. For every omitted index $j>n$, we have $s_j\le s_{n+1}$, and hence \begin{align*} \|(T-T_n)x\|_K^2 \le s_{n+1}^2\sum_{\substack{j\in J, j>n}} |(x,v_j)_H|^2. \end{align*} Since $(v_j)_{j\in J}$ is an orthonormal set in $H$, [Bessel's inequality](/theorems/540) gives \begin{align*} \sum_{\substack{j\in J, j>n}} |(x,v_j)_H|^2\le \|x\|_H^2. \end{align*} Combining the last two displayed formulas yields \begin{align*} \|(T-T_n)x\|_K\le s_{n+1}\|x\|_H. \end{align*} Taking the supremum over all $x\in H$ with $\|x\|_H\le 1$ proves \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)}\le s_{n+1}. \end{align*} To see that this upper bound is sharp, we test the tail on the first omitted right singular vector. The vector $v_{n+1}$ has norm $1$, and all coefficients $(v_{n+1},v_j)_H$ vanish except the coefficient with $j=n+1$. Therefore \begin{align*} (T-T_n)v_{n+1}=s_{n+1}u_{n+1}. \end{align*} Since $u_{n+1}$ is also a unit vector in $K$, \begin{align*} \|(T-T_n)v_{n+1}\|_K=s_{n+1}. \end{align*} The definition of the operator norm then gives \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)} \ge \|(T-T_n)v_{n+1}\|_K = s_{n+1}. \end{align*} Together with the upper bound, this proves \begin{align*} \|T-T_n\|_{\mathcal{L}(H,K)}=s_{n+1}. \end{align*} If there are at most $n$ non-zero singular values, then no tail remains after the truncation: $T_n=T$. In that case the convention gives $s_{n+1}=0$, and the equality becomes $\|T-T_n\|_{\mathcal{L}(H,K)}=0$.[/guided]

[step:Find a unit vector killed by any rank-at-most-$n$ competitor] Let $R\in\mathcal{L}(H,K)$ satisfy $\operatorname{rank}R\le n$. If $s_{n+1}=0$, then the desired inequality \begin{align*} \|T-R\|_{\mathcal{L}(H,K)}\ge 0 \end{align*} is immediate. We therefore assume $s_{n+1}>0$, so $J$ contains $1,\dots,n+1$. Define the finite-dimensional subspace $V_{n+1}\subset H$ by \begin{align*} V_{n+1}:=\operatorname{span}\{v_1,\dots,v_{n+1}\}. \end{align*} Since the vectors $v_1,\dots,v_{n+1}$ are orthonormal, $\dim V_{n+1}=n+1$. Consider the restricted [linear map](/page/Linear%20Map) \begin{align*} R|_{V_{n+1}}:V_{n+1}\to \operatorname{Range}(R). \end{align*} Its range has dimension at most $n$. Hence its kernel has positive dimension: \begin{align*} \dim\ker(R|_{V_{n+1}}) = \dim V_{n+1}-\dim \operatorname{Range}(R|_{V_{n+1}}) \ge (n+1)-n = 1. \end{align*} Choose a non-zero vector $y\in\ker(R|_{V_{n+1}})$, and define \begin{align*} x:=\frac{y}{\|y\|_H}. \end{align*} Then $x\in V_{n+1}$, $\|x\|_H=1$, and $Rx=0$. [/step]

[step:Use the singular-value ordering on this vector to force the lower bound] Because $x\in V_{n+1}$ and $(v_1,\dots,v_{n+1})$ is an [orthonormal basis](/page/Orthonormal%20Basis) of $V_{n+1}$, we have \begin{align*} x=\sum_{j=1}^{n+1}(x,v_j)_H v_j. \end{align*} Applying the singular-value expansion to this finite linear combination gives \begin{align*} Tx=\sum_{j=1}^{n+1}s_j(x,v_j)_H u_j. \end{align*} Using the orthonormality of $(u_1,\dots,u_{n+1})$ in $K$, \begin{align*} \|Tx\|_K^2 = \sum_{j=1}^{n+1}s_j^2 |(x,v_j)_H|^2. \end{align*} For $1\le j\le n+1$, the non-increasing order of the singular values gives $s_j\ge s_{n+1}$. Therefore \begin{align*} \|Tx\|_K^2 \ge s_{n+1}^2\sum_{j=1}^{n+1}|(x,v_j)_H|^2 = s_{n+1}^2\|x\|_H^2 = s_{n+1}^2. \end{align*} Since $Rx=0$ and $\|x\|_H=1$, the definition of the operator norm gives \begin{align*} \|T-R\|_{\mathcal{L}(H,K)} \ge \|(T-R)x\|_K = \|Tx\|_K \ge s_{n+1}. \end{align*} This proves that every rank-at-most-$n$ operator $R$ satisfies \begin{align*} \|T-R\|_{\mathcal{L}(H,K)}\ge s_{n+1}. \end{align*} Together with the equality for the Schmidt truncation, this completes the proof. [/step]